Em 2

1 CHAPTER 2 ELECTROSTATIC POTENTIAL 2.1 Introduction Imagine that some region of space, such as the room you are sitting in, is permeated by an electric field. (Perhaps there are all sorts of electrically charged bodies outside the room.) If you place a small positive test charge somewhere in the room, it will experience a force F = QE. If you try to move the charge from point A to point B against the direction of the electric field, you will have to do work. If work is required to move a positive charge from point A to point B, there is said to be an electrical potential difference between A and B, with point A being at the lower potential. If one joule of work is required to move one coulomb of charge from A to B, the potential difference between A and B is one volt (V). The dimensions of potential difference are ML2T−2Q−1. All we have done so far is to define the potential difference between two points. We cannot define “the” potential at a point unless we arbitrarily assign some reference point as having a defined potential. It is not always necessary to do this, since we are often interested only in the potential differences between point, but in many circumstances it is customary to define the potential to be zero at an infinite distance from any charges of interest. We can then say what “the” potential is at some nearby point. Potential and potential difference are scalar quantities. Suppose we have an electric field E in the positive x-direction (towards the right). This means that potential is decreasing to the right. You would have to do work to move a positive test charge Q to the left, so that potential is increasing towards the left. The force on Q is QE, so the work you would have to do to move it a distance dx to the right is −QE dx, but by definition this is also equal to Q dV, where dV is the potential difference between x and x + dx. Therefore

E = −

dV . dx

2.1.1

In a more general three-dimensional situation, this is written ∂V ∂V   ∂V +k E = − grad V = − ∇ V = −  i +j . ∂x ∂x   ∂x

2.1.2

We see that, as an alternative to expressing electric field strength in newtons per coulomb, we can equally well express it is volts per metre (V m−1). The inverse of equation 2.1.1 is, of course,

2 V = − ∫ E dx + constant .

2.1.3

2.2 Potential Near Various Charged Bodies 2.2.1 Point Charge Let us arbitrarily assign the value zero to the potential at an infinite distance from a point charge Q. “The” potential at a distance r from this charge is then the work required to move a unit positive charge from infinity to a distance r. At a distance x from the charge, the field strength is

Q . The work required to 4πε0 x 2

Q δx . The work required to move unit 4πε0 x 2 ∞ dx Q Q . charge from r to infinity is − The work required to move unit = − 2 ∫ 4πε0 r x 4πε0 r charge from infinity to r is minus this.

move a unit charge from x to x + δx is −

V = +

Therefore

Q . 4πε0 r

2.2.1

The mutual potential energy of two charges Q1 and Q2 separated by a distance r is the work required to bring them to this distance apart from an original infinite separation. This is QQ P.E. = + 1 2 . 2.2.2 4πε0 r Before proceeding, a little review is in order. Field at a distance r from a charge Q: Q , 4πε0 r 2

N C−1 or V m−1

Q Q rˆ = r. 2 4πε0 r 4πε0 r 3

N C−1 or V m−1

E =

or, in vector form,

E =

3 Force between two charges, Q1 and Q2: F =

Q1Q2 . 4πε r 2

N

Potential at a distance r from a charge Q: V =

Q . 4πε0 r

V

Mutual potential energy between two charges:

P.E. =

Q1Q2 . 4πε0 r

J

We couldn’t possibly go wrong with any of these, could we?

2.2.2

Spherical Charge Distributions

Outside any spherically-symmetric charge distribution, the field is the same as if all the charge were concentrated at a point in the centre, and so, then, is the potential. Thus V =

Q . 4πε0 r

2.2.3

Inside a hollow spherical shell of radius a and carrying a charge Q the field is zero, and therefore the potential is uniform throughout the interior, and equal to the potential on the surface, which is V =

Q . 4πε 0 a

2.2.4

A solid sphere of radius a bearing a charge Q that is uniformly distributed throughout the sphere is easier to imagine than to achieve in practice, but, for all we know, a proton might be like this (it might be – but it isn’t!), so let’s calculate the field at a point P inside the sphere at a distance r (< a) from the centre. See figure II.1 We can do this in two parts. First the potential from the part of the sphere “below” P. If Qr . the charge is uniformly distributed throughout the sphere, this is just Here Qr is 4πε0 r

4 the charge contained within radius r, which, if the charge is uniformly distributed Q r2 . throughout the sphere, is Q(r 3 / a 3 ) . Thus, that part of the potential is 4πε0 a 3

P r

FIGURE II.1 Next, we calculate the contribution to the potential from the charge “above” P. Consider an elemental shell of radii x , x + δx. The charge held by it is 2 2 4π x δx 3Qx δx . δQ = 4 3 × Q = The contribution to the potential at P from the charge a3 3 πa δQ 3Q x δx . in this elemental shell is = The contribution to the potential from all 4π ε 0 x 4πε 0 a 3 3Q 4πε0 a 3 of the potential, we obtain the charge “above” P is

∫

a r

x dx =

3Q(a 2 − r 2 ) . Adding together the two parts 4πε0 .2 a 3

5 V =

Q (3a 2 − r 2 ) . 3 8πε 0 a

2.2.5

2.2.3 Long Charged Rod The field at a distance r from a long charged rod carrying a charge λ coulombs per metre λ . Therefore the potential difference between two points at distances a and b is 2πε0 r from the rod (a < b) is Vb − Va = −

â

Va − Vb =

λ 2πε 0

∫

b

a

dr . r

λ ln(b / a ) . 2πε0

2.2.6

2.2.4 Large Plane Charged Sheet The field at a distance r from a large charged sheet carrying a charge σ coulombs per σ . Therefore the potential difference between two points at distances square metre is 2ε 0 a and b from the sheet (a < b) is Va − Vb =

σ (b − a ) . 2ε 0

2.2.7

2.2.5 Axis of a Charged Ring The field on the axis of a charged ring is given in section 1.6.4. The reader is invited to show that the potential on the axis of the ring is V =

Q . 4πε 0 (a + x 2 )1/ 2 2

2.2.8

You can do this either by integrating the expression for the field or just by thinking about it for a few seconds and realizing that potential is a scalar quantity.

6 2.2.6 Axis of a Charged Disc The field on the axis of a charged disc is given in section 1.6.5. The reader is invited to show that the potential on the axis of the ring is V =

2Q [(a 2 + x 2 )1/ 2 − x] . 2 4πε 0 a

2.2.9

2.3 Electron-volts The electron-volt is a unit of energy or work. An electron-volt (eV) is the work required to move an electron through a potential difference of one volt. Alternatively, an electronvolt is equal to the kinetic energy acquired by an electron when it is accelerated through a potential difference of one volt. Since the magnitude of the charge of an electron is about 1.602 × 10−19 C, it follows that an electron-volt is about 1.602 × 10−19 J. Note also that, because the charge on an electron is negative, it requires work to move an electron from a point of high potential to a point of low potential. Exercise. If an electron is accelerated through a potential difference of a million volts, its kinetic energy is, of course, 1 MeV. At what speed is it then moving? First attempt.

1 2

mv 2 = eV .

(Here eV, written in italics, is not intended to mean the unit electron-volt, but e is the magnitude of the electron charge, and V is the potential difference (106 volts) through which it is accelerated.) Thus v = 2eV / m . With m = 9.109 × 10−31 kg, this comes to v = 5.9 × 108 m s−1. Oops! That looks awfully fast! We’d better do it properly this time. Second attempt.

( γ − 1)mc 2 = eV .

Some readers will know exactly what we are doing here, without explanation. Others may be completely mystified. For the latter, the difficulty is that the speed that we had calculated was even greater than the speed of light. To do this properly we have to use the formulas of special relativity. See, for example, Chapter 15 of the Classical Mechanics section of these notes. At any rate, this results in γ = 2.958, whence β = 0.9411 and v = 2.82 × 108 m s−1.

7 2.4 A Point Charge and an Infinite Conducting Plane An infinite plane metal plate is in the xy-plane. A point charge +Q is placed on the z-axis at a height h above the plate. Consequently, electrons will be attracted to the part of the plate immediately below the charge, so that the plate will carry a negative charge density σ which is greatest at the origin and which falls off with distance ρ from the origin. Can we determine σ(ρ)? See figure II.2

+Q

h

−

−

−

−

−

− − − − −−− − − − −

σ −

−

−

−

−

ρ FIGURE II.2

First, note that the metal surface, being a conductor, is an equipotential surface, as is any metal surface. The potential is uniform anywhere on the surface. Now suppose that, instead of the metal surface, we had (in addition to the charge +Q at a height h above the xy-plane), a second point charge, −Q, at a distance h below the xy-plane. The potential in the xy-plane would, by symmetry, be uniform everywhere. That is to say that the potential in the xy-plane is the same as it was in the case of the single point charge and the metal plate, and indeed the potential at any point above the plane is the same in both cases. For the purpose of calculating the potential, we can replace the metal plate by an image of the point charge. It is easy to calculate the potential at a point (z , ρ). If we suppose that the permiittivity above the plate is ε0, the potential at (z , ρ) is

V =

 Q  1 1  2 . − 2 2 1/ 2 2 1/ 2  4πε 0  [ρ + (h − z ) ] [ρ + (h + z ) ] 

2.4.1

8 The field strength E in the xy-plane is −∂V / ∂z evaluated at z = 0, and this is E = −

2Q . h . 2 4 πε 0 (ρ + h 2 ) 3/ 2

2.4.2

The D-field is ε0 times this, and since all the lines of force are above the metal plate, Gauss's theorem provides that the charge density is σ = D, and hence the charge density is

This can also be written

σ = −

Q. h . 2 2 π ( ρ + h 2 ) 3/ 2

2.4.3

σ = −

Q. h , 2π ξ3

2.4.4

where ξ 2 = ρ2 + h 2 , with obvious geometric interpretation. Exercise: How much charge is there on the surface of the plate within an annulus bounded by radii ρ and ρ + dρ? Integrate this from zero to infinity to show that the total charge induced on the plate is −Q. 2.5 A Point Charge and a Conducting Sphere

P a

O

ξ

ζ

I

R

+Q

FIGURE II.3

A point charge +Q is at a distance R from a metal sphere of radius a. We are going to try to calculate the surface charge density induced on the surface of the sphere, as a function of position on the surface. We shall bear in mind that the surface of the sphere is an equipotential surface, and we shall take the potential on the surface to be zero.

9 Let us first construct a point I such that the triangles OPI and PQO are similar, with the lengths shown in figure II.3. The length OI is a2/R. Then R / ξ = a / ζ , or 1 a/R − = 0. ξ ζ

2.5.1

This relation between the variables ξ and ζ is in effect the equation to the sphere expressed in these variables. Now suppose that, instead of the metal sphere, we had (in addition to the charge +Q at a distance R from O), a second point charge −(a/R)Q at I. The locus of points where the potential is zero is where Q  1 a /R   = 0.  − 4πε0  ξ ζ 

2.5.2

That is, the surface of our sphere. Thus, for purposes of calculating the potential, we can replace the metal sphere by an image of Q at I, this image carrying a charge of −(a/R)Q. Let us take the line OQ as the z-axis of a coordinate system. Let X be some point such that OX = r and the angle XOQ = θ. The potential at P from a charge +Q at Q and a charge −(a/R)Q at I is (see figure II.4)

X r

+Q

O

I OI = a2/R

FIGURE II.4

V =

Q 4πε0

  a /R 1  2 2 . − 2 4 2 1/ 2 2 1/ 2  (r + a / R −2a r cos θ /R)   (r + R −2rR cos θ)

2.5.2

10 The E field on the surface of the sphere is −∂V / ∂r evaluated at r = a. The D field is ε0 times this, and the surface charge density is equal to D. After some patience and algebra, we obtain, for a point X on the surface of the sphere

σ = −

Q . R 2 −a 2 . 1 . 4π a (XQ) 3

2.5.3

2.6 Two Semicylindrical Electrodes This section requires that the reader should be familiar with functions of a complex variable and conformal transformations. For readers not familiar with these, this section can be skipped without prejudice to understanding following chapters. For readers who are familiar, this is a nice example of conformal transformations to solve a physical problem.

y

V = 0 A B

O V = V0

FIGURE II.5

We have two semicylindrical electrodes as shown in figure II.5. The potential of the upper one is 0 and the potential of the lower one is V0. We'll suppose the radius of the curcle is 1; or, what amounts to the same thing, we'll express coordinates x and y in units

x

11 of the radius. Let us represent the position of any point whose coordinates are (x , y) by a complex number z = x + iy. 1 − z   ; that is, Now let w = u + iv be a complex number related to z by w = i 1 + z  1 + iw . Substitute w = u + iv and z = x + iy in each of these equations, and equate z= 1 − iw real and imaginary parts, to obtain u =

2y ; (1 + x ) 2 + y 2

v =

1 − x2 − y2 ; (1 + x ) 2 + y 2

2.6.1

x =

1 − u2 − v 2 ; u 2 + (1 + v ) 2

y =

2u . u + (1 + v ) 2

2.6.2

2

In that case, the upper semicircle (V = 0) in the xy-plane maps on to the positive u-axis in the uv-plane, and the lower semicircle (V = V0) in the xy-plane maps on to the negative u-axis in the uv-plane. (Figure II.6.) Points inside the circle bounded by the electrodes in the xy-plane map on to points above the u-axis in the uv-plane.

v

B

V = V0

O

V = 0

u

FIGURE II.6

In the uv-plane, the lines of force are semicircles, such as the one shown. The potential goes from 0 at one end of the semicircle to V0 at the other, and so equation to the semicircular line of force is arg w V = π V0

2.6.3

A

12

V =

or

V0 tan −1 (v / u). π

2.6.4

The equipotentials (V = constant) are straight lines in the uv-plane of the form v = fu.

2.6.5

(You would prefer me to use the symbol m for the slope of the equipotentials, but in a moment you will be glad that I chose the symbol f.) If we now transform back to the xy-plane, we see that the equation to the lines of force is V =

 1 − x2 − y 2  V0  , tan −1  2y π  

2.6.6

and the equation to the equipotentials is

or

1 − x 2 − y 2 = 2 fy,

2.6.7

x 2 + y 2 + 2 fy − 1 = 0.

2.6.8

Now aren't you glad that I chose f ? Those who are handy with conic sections (see Chapter 2 of Celestial Mechanics) will understand that the equipotentials in the xy-plane are circles of radii f 2 + 1 , whose centres are at (0 , ! f ), and which all pass through the points (!1 , 0). They are drawn as blue lines in figure II.7. The lines of force are the orthogonal trajectories to these, and are of the form x 2 + y 2 + 2 gy + 1 = 0.

2.6.9

These are circles of radii g 2 − 1 and have their centres at (0 , ! g). They are shown as dashed red lines in figure II.7.

13

FIGURE II.7